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For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is $p$. If he fails in one of the exams then the probability of passing in the next exam is $p/2$ otherwise it remains the same. Find the probability that he will qualify. My textbook answer reads $2p^2 – p^3$. This is possible if only the below cases are considered:

  1. He passes first and second exam.
  2. He passes first, fails in second but passes third exam.
  3. He fails in first, passes second and third exam.

But I think this is wrong since at least two out of three exams means,passing in first, second and third exam is inclusive. Someone please solve this paradox.

N. F. Taussig
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sree_ranjani
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5 Answers5

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See the probability tree diagram: enter image description here

Adding the qualifying probabilities we get: $$p^3+p^2(1-p)+p(1-p)\cdot \frac p2+(1-p)\cdot \frac p2 \cdot p=2p^2-p^3.$$

farruhota
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I get: p^2 + (1-p)(p/2)p + p(1-p)(p/2)

Which is (pass pass either) + (fail pass pass) + (pass fail pass), and as you can see no cases are double counted here. If you then expand out the (1-p) terms and collect it together then it's the same as the textbook answer.

Steven Irrgang
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  • all your cases are for not for atleast 2,but for just 2. – sree_ranjani Apr 26 '18 at 08:49
  • @parker no they're not, the case of all 3 is included in (pass pass either). If it makes it easier, split that case into (pass pass pass) and (pass pass fail), and change p^2 to p^3 + p^2(1-p) – Steven Irrgang Apr 26 '18 at 08:51
  • no still not clear.can you explain (pass pass pass) and (pass pass fail)? – sree_ranjani Apr 26 '18 at 08:54
  • @parker (pass pass pass) = pass all three exams (pass pass fail) means pass the first two exams and fail the third. – Steven Irrgang Apr 26 '18 at 08:59
  • Then the case is (pass pass pass) + (fail pass pass) + (pass fail pass) . this won't give the same answer .The case (pass pass fail)is rejected cause its the same as (fail pass pass). – sree_ranjani Apr 26 '18 at 09:02
  • @parker You need to consider all sequences in which there are at least two passes. They are $(\text{pass}, \text{pass}, \text{pass})$, $(\text{pass}, \text{pass}, \text{fail})$, $(\text{pass}, \text{fail}, \text{pass})$, $(\text{fail}, \text{pass}, \text{pass})$. The first two cases have a combined probability of $ppp + pp(1 - p) = pp(p + 1 - p) = p^2$. The third case has probability $p(1 - p)\frac{p}{2}$. The fourth case has probability has probability $(1 - p)\frac{p}{2}p$. – N. F. Taussig Apr 26 '18 at 11:21
  • That seems the correct reason. Thanks @N.F.Taussig – sree_ranjani Apr 26 '18 at 14:30
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PP has probability $p^2$.

PFP has probability $p(1-p)\frac12p$.

FPP has probability $(1-p)\frac12pp$

Summation results in a probability of $2p^2-p^3$ to qualify.

drhab
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All three passed: $p^3$. First and second passed, third failed: $p^2(1-p)$. First passed, second failed, third passed: $p(1-p)p/2$. First failed, next two passed: $(1-p)(p/2)^2$. Then add all four probabilities. The answer would be $7p^2/4-3p^3/4$.

UPDATE. I understood it so that if once failed, the probability of success remains at most $p/2$. But I suppose this is not correct. So the case "First failed, next two passed" leads to $(1-p)(p/2)p$ and the whole probability is indeed $2p^2-p^3$. The problem is badly stated, by the way.

Logic_Problem_42
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Let me denote by $E_i$ the event "Exam $i$ has been passed" and let $\overline{E}_i$ be its complement. Then, you know that $P(E_1)=p$, $P(E_2| \overline{E}_1 ) = P(E_3| \overline{E}_1 \cap \overline{E}_2 ) = P(E_3| \overline{E}_1 \cap E_2 ) = P(E_3| E_1 \cap \overline{E}_2 ) = p/2$ and $P(E_2| E_1 ) = P(E_3| E_1 \cap E_2 ) = p$.

By $\sigma$-addivity, you want to calculate the probability $$ A=P(E_1\cap E_2 \cap E_3) + P(E_1\cap E_2\cap \overline{E}_3) + P(E_1\cap \overline{E}_2 \cap E_3) + P(\overline{E}_1 \cap E_2\cap E_3).$$ Let's calculate each term: $$ P(E_1\cap E_2 \cap E_3) = P(E_3 | E_1\cap E_2) P(E_1\cap E_2) = p P( E_2 | E_1) P(E_1) = p^3 $$ $$ P(E_1\cap E_2\cap \overline{E}_3) = P(\overline{E}_3 | E_1\cap E_2) P( E_2 | E_1) P(E_1) = (1-p) p^2 $$ $$ P(E_1\cap \overline{E}_2 \cap E_3) = P(E_3 | E_1\cap \overline{E}_2) P(E_1\cap \overline{E}_2) = \frac{p}{2} P(\overline{E}_2 | E_1) P(E_1) = \frac{p}{2} (1-p) p $$ $$ P(\overline{E}_1 \cap E_2\cap E_3) = P(E_3 | \overline{E}_1 \cap E_2 ) P (\overline{E}_1 \cap E_2) = \frac{p}{2} P(E_2|\overline{E}_1) P(\overline{E}_1) = \frac{p^2}{4}(1-p) $$ To sum up, we obtain $$ A = p^3 + (1-p) p^2 + \frac{p}{2} (1-p) p + \frac{p^2}{4}(1-p) % = p^3 + p^2- p^3 + \frac{p^2-p^3}{2} + \frac{p^2}{4} - \frac{p^3}{4} % = p^2 + 3 p^2 \frac{1-p}{4} % = p^2 + \frac{3 p^2}{4} - \frac{3 p^3}{4} = \frac{7 }{4}p^2 - \frac{3 }{4} p^3 $$

user52227
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